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LT1107

1107fa

Inductor Selection –– Step-Up Converter

In a step-up, or boost converter (Figure 1), power gener-

ated by the inductor makes up the difference between

input and output. Power required from the inductor is

determined by:

where VD is the diode drop (0.5V for a 1N5818 Schottky).

Energy required by the inductor per cycle must be equal or

greater than:

in order for the converter to regulate the output.

When the switch is closed, current in the inductor builds

according to:

where R

′ is the sum of the switch equivalent resistance

(0.8

Ω typical at 25°C) and the inductor DC resistance.

When the drop across the switch is small compared to VIN,

the simple lossless equation:

As an example, suppose 12V at 60mA is to be generated

from a 3V to 6V input. Recalling equation (1),

Energy required from the inductor is:

Picking an inductor value of 33

µH with 0.2Ω DCR results

in a peak switch current of:

Substituting IPEAK into Equation 4 results in:

Since 11.9

µJ > 9.05µJ, the 33µH inductor will work. This

trial-and-error approach can be used to select the opti-

mum inductor.

A resistor can be added in series with the ILIM pin to invoke

switch current limit. The resistor should be picked so the

calculated IPEAK at minimum VIN is equal to the Maximum

Switch Current (from Typical Performance Characteristic

curves). Then, as VIN increases, peak switch current is

held constant, resulting in increasing efficiency.

Inductor Selection –– Step-Down Converter

The step-down case (Figure 2) differs from the step-up in

that the inductor current flows through the load during

both the charge and discharge periods of the inductor.

Current through the switch should be limited to ~650mA

in this mode. Higher current can be obtained by using an

external switch (see LT1111 and LT1110 data sheets). The

ILIM pin is the key to successful operation over varying

inputs.

After establishing output voltage, output current and input

voltage range, peak switch current can be calculated by the

formula:

APPLICATI

I FOR ATIO

OSC

/( )

()

( )

′

−













− ′

()=

()

can be used. These equations assume that at t = 0,

inductor current is zero. This situation is called “discon-

tinuous mode operation” in switching regulator parlance.

Setting “t” to the switch ON time from the LT1107 speci-

fication table (typically 11

µs) will yield IPEAK for a specific

“L” and VIN. Once IPEAK is known, energy in the inductor

at the end of the switch ON time can be calculated as:

EL must be greater than PL/fOSC for the converter to deliver

the required power. For best efficiency IPEAK should be

kept to 1A or less. Higher switch currents will cause

excessive drop across the switch resulting in reduced

efficiency. In general, switch current should be held to as

low a value as possible in order to keep switch, diode and

inductor losses at a minimum.

kHz

OSC

570

905

.( )

L =

()( ) =

085

1191

µµ

( )

ELI

PEAK

()

L =+

−

()( )=

0 5

570

.( )

emA

PEAK

=−













 =

−

850

111

Ω

Ω•

()

OUT

IN MIN

OUT

−







 ()

()

PEAK

OUT

−+













()

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